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TH
0
N
22 tháng 10 2018
\(x^2-11x-26=0\)
\(x^2-13x+2x-26=0\)
\(x.\left(x-13\right)+2.\left(x-13\right)=0\)
\(\left(x+2\right).\left(x-13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-13=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=13\end{cases}}\)
vậy...
P/S: lớp 7 sai sót mong thông cảm
TN
1 tháng 7 2017
\(A=x^2-10x+26\)
\(=\left(x^2-10x+25\right)+1\)
\(=\left(x-5\right)^2+1\ge1\)
Vậy \(Min_A=1\) khi \(x-5=0\Rightarrow x=5\)
\(B=x^2+7x+10=\left(x^2+7x+\dfrac{49}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{7}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)Vậy \(Min_B=\dfrac{-9}{4}\) khi \(x+\dfrac{7}{2}=0\Rightarrow x=\dfrac{-7}{2}\)
\(C=4x^2+8x+15=4\left(x^2+2x+1\right)+11=4\left(x+1\right)^2+11\ge11\)Vậy \(Min_C=11\) khi \(x+1=0\Rightarrow x=-1\)
\(D=3x^2-7x+20=3\left(x^2-\dfrac{7}{3}x+\dfrac{49}{36}\right)+\dfrac{191}{12}=3\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{12}\ge\dfrac{191}{12}\)Vậy \(Min_D=\dfrac{191}{12}\) khi \(x-\dfrac{7}{6}=0\Rightarrow x=\dfrac{7}{6}\)
\(E=x^2-4xy+5y^2-22y+8\)
\(=\left(x^2-4xy+4y^2\right)+\left(y^2-22y+121\right)-113\)\(=\left(x-2y\right)^2+\left(y-11\right)^2-113\ge-113\)
Vậy \(Min_E=-113\) khi \(\left[{}\begin{matrix}x-2y=0\\x-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}11-2y=0\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2y=11\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=11\end{matrix}\right.\)
KB
22 tháng 10 2018
a ) \(x^2-11x-26=0\)
\(\Leftrightarrow x^2-13x+2x-26=0\)
\(\Leftrightarrow x\left(x-13\right)+2\left(x-13\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=13\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-2\\x=13\end{matrix}\right.\)
b ) \(2x^2+7x-4=0\)
\(\Leftrightarrow2\left(x^2+\dfrac{7}{2}x-2\right)=0\)
\(\Leftrightarrow x^2+\dfrac{7}{2}x-2=0\)
\(\Leftrightarrow x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{81}{16}=0\)
\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2=\dfrac{81}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{9}{4}\\x+\dfrac{7}{4}=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-4\end{matrix}\right.\)
c ) \(\left(x-2\right)\left(x-3\right)+\left(x-2\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
26 tháng 6 2023
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
ND
0
15 tháng 7 2021
a) \(\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
<=> \(9x^2-9x+2=9x^2+6x+1\)
<=> \(15x=1\) <=> \(x=\frac{1}{15}\)
b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)
<=> \(4x^2+3x-1=4x^2-12x+9\)
<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)
c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)
<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)
<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)
15 tháng 7 2021
d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)
<=> 16 - 9x2 = 12x - 9x2 - 3
<=> 12x = 19
<=> x = 19/12
e) x(x + 1)(x + 2)(x + 3) = 24
<=> (x2 + 3x)(x2 + 3x + 2) = 24
<=> (x2 + 3x)2 + 2(x2 + 3x) - 24 = 0
<=> (x2 + 3x)2 + 6(x2 + 3x) - 4(x2 + 3x) - 24 = 0
<=> (x2 + 3x + 6)(x2 + 3x - 4) = 0
<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)
g) (7x - 2)2 = (7x - 3)(7x + 2)
<=> 49x2 - 28x + 4 = 49x2 - 7x - 6
<=> 21x = 10 <=> x = 10/21
TT
11 tháng 10 2020
Đi học về mở máy ra thấy đống câu chưa trả lời :vv câu đầu tiên của buổi :33
a, \(x^3-9x^2+6x+16\)
\(=\left(x^3+x^2\right)+\left(-10x^2-10x\right)+\left(16x+16\right)\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[\left(x^2-2x\right)+\left(-8x+16\right)\right]\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)
b, \(x^2+2x-24\)
\(=\left(x^2+4x\right)-\left(6x+24\right)\)
\(=x\left(x+4\right)-6\left(x+4\right)\)
\(=\left(x-6\right)\left(x+4\right)\)
c, \(3x^2+7x-26\)
\(=\left(3x^2-6x\right)+\left(13x-26\right)\)
\(=3x\left(x-2\right)+13\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+13\right)\)
=>21x-78=2x-2
=>19x=76
=>x=4
\(\dfrac{7x-26}{x-1}=\dfrac{2}{3}đkxđ\left(x\ne1\right)\\ \Leftrightarrow\dfrac{3\left(7x-26\right)}{3\left(x-1\right)}=\dfrac{2\left(x-1\right)}{3\left(x-1\right)}\\ \Rightarrow3\left(7x-26\right)=2\left(x-1\right)\\ \Leftrightarrow21x-78=2x-2\\ \Leftrightarrow21x-2x=-2+78\\ \Leftrightarrow19x=76\\ \Leftrightarrow x=4\)
Vậy pt có nghiệm `x=4`